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When going through one of my chemistry textbooks, I saw that the electrolysis of aqueous lead nitrate led to oxygen being formed at the anode and lead being formed at the cathode.

However, in school, I was taught that only metals less reactive than hydrogen will form at the cathode. This electrolysis contradicts what I learned, as lead is more reactive than hydrogen, and should remain as ions in solution.

Therefore, my question is: What forms at the cathode during electrolysis of aqueous lead nitrate and why?

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Firstly, the reactivity being referred to is not reactivity in the sense of the reactivity series of metals, but rather in the sense of standard electrode potentials (standard reduction potentials).

This comes into play when setting up an electrolytic cell in determining the reaction that will take place at the cathode/anode. In this case, the reaction that requires the least amount of work (or voltage in this case, since voltage is essentially a measure of work) will take place. So, assuming that this is done aqueously, if the reduction/oxidation reaction to take place requires more voltage than the reduction/oxidation of the water, then the water will simply be reduced/oxidized instead. (I think this is where the reactivity of hydrogen comes into play in how you were taught).

So to answer your question more directly, it is actually lead that forms at the cathode.

Edit: To add some detail, the reduction potential of H2O is -0.8277 V, while the lead(II) ion is -0.13. In this case, less energy means closer to zero, the negative being essentially irrelevant. Hope this helped.

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The sentence learnt at school is not perfectly correct. It should not be :

In school, I was taught that only metals less reactive than hydrogen will form at the cathode.

It should have been modified by adding "at comparable ionic concentration", to give :

In school, I was taught that only metals less reactive than hydrogen at comparable ionic concentration will form at the cathode.

As an example, in $1 M$ solution of $\ce{Pb^{2+}}$ or $\ce{Sn^{2+}}$ or $\ce{Ni^{2+}}$, containing also $\ce{H+}$ with the same concentration $\ce{[H+]} = 1 M$, the metals $\ce{Pb, Sn, Ni}$ will not be deposited by electrolysis. $\ce{H2}$ will be deposited before the metals, as their redox potentials are too negative, being respectively $-0.13 V, -0.14 V$, and $-0.23 V$, and because of the redox potential of $\ce{H2/H+} = 0 V$

But in non-acidic solution, the situation is reversed. In the same $1 M$ solutions of $\ce{Pb^{2+}, Sn^{2+}, Ni^{2+}}$ but with $\ce{[H+]}$ = $\ce{10^{-7}}M$, the redox potential of $\ce{H2/H+}$ becomes very negative, being = $-0.41 V$. The metals are then deposited by electrolysis before $\ce{H2}$, because at $p$H $7$, all three redox potentials are less negative than $-0.41 V$

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What was stated is correct but there is an exception. For the metals, zinc to lead in the reactivity series( zn, Fe, Pb and what's in between) what is formed at the cathode can be affected by concentration. If the solution is concentrated, the metal forms at the cathode and if the solution is dilute, hydrogen forms. Hope this helps.

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